Post on 09-Jan-2022
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ECE 255: L17
BJT Transistor Amplifiers (Sedra and Smith, 7th Ed., Sec. 7.1, 7.2.2)
Mark Lundstrom School of ECE
Purdue University West Lafayette, IN USA
Spring 2019 Purdue University
Lundstrom: 2019
Announcements
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1) LTspice Project 2 is posted. The due date for it is Wed. Feb. 27th by 5:00PM electronically. You should find your assigned your beta value in your grade book.
2) HW6 will be posted today
3) Exam 2 in on Tuesday, March 5, 6:30-7:30 PM PHYS 112
Lundstrom: 2019
Outline
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1) Voltage transfer characteristic (VTC) 2) Small signal model for BJT 3) Small signal analysis
Lundstrom: 2019
Voltage Transfer Characteristic
4
+V
OUT=VCC − ICRC
−
β = 100
VCC = 10 V
RC = 5 kΩIC
RBB = 430 kΩ
VIN+−
VBE ON( ) = 0.7 V
What is VOUT vs. VIN?
Lundstrom: 2019
VTC
5
VIN
VOUT
VCC
≈ 0.3V
≈ 0.7 V
cutoff active saturation
VOUT
VCC = 10 V
RC = 5 kΩIC
RBB = 430 kΩ
VIN+−
Lundstrom: 2019
Biasing in the active region
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VIN
VOUT
VCC
≈ 0.3V
≈ 0.7 V
cutoff active saturation
Q : IC ,VCE( )
VOUT =VCC − IC RC
VOUT =VCC − ISeVIN VT( )RC
Lundstrom: 2019
Bias Circuit
7
+V
CE
−
β = 100
VCC = 10 V
RC = 5 kΩIC
RBB = 430 kΩ
VBB = 5 V +−
IB = ? mA
IC = ? mA
VCE = ? V
VBE ON( ) = 0.7 V Lundstrom: 2019
Bias Circuit
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+V
CE
−
β = 100
VCC = 10 V
RC = 5 kΩIC
RBB = 430 kΩ
VBB = 5 V +−
IB = 0.01mA
IC = 1.0 mA
VCE = 5.0 V
VBE ON( ) = 0.7 V Lundstrom: 2019
Biasing
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VIN
VOUTVCC = 10 V
≈ 0.3V
cutoff active saturation
Q : IC = 1mA,VCE = 5.0 V( )
VO = 5 V
VBB = 5.0 V
Lundstrom: 2019
Add ac small signal
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+υO =V
O+υo sinωt
−
β = 100
VCC = 10 V
RC = 5 kΩiC = IC + ic
RBB = 430 kΩ
VBB = 5 V+−
υi sinωt
VBE ON( ) = 0.7 V
“Common emitter amplifier”
Lundstrom: 2019
Qualitative
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VI
VO
VCCAV = δVO
δVI≈ dVOdVI
AV < 0
voltage gain
υ I =VBB +υiυ I =VBB −υi
VO
VO −δVO
VO +δVO
(for this circuit)
2δVI
2δVO
Qualitative
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VI
VO
VCC
≈ 0.3V
VO =VCC − IC RC
VO =VCC − ISeVBE VT( )RC
dVO
dVBE
= −ISeVBE VT( )
VT
RC = −IC
VT
RC
gm ≡
dIC
dVBE
=IC
VT(transconductance)
Aυbe=υo
υbe
= −gmRC
Lundstrom: 2019
Voltage gain of CE amplifier
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+υOUT =VO
+υo
−
β = 100
VCC = 10 V
RC = 5 kΩiC = IC + ic
RBB = 430 kΩ
VBB = 5 V +−
υi
VBE ON( ) = 0.7 V
+υbe
−
Aυbe
=υo
υbe
= −gmRC
Aυi
=υo
υi
= ?
Lundstrom: 2019
Analysis by superposition
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+υOUT =VO
+υo
−
β = 100
VCC = 10 V
RC = 5 kΩiC = IC + ic
RBB = 430 kΩ
VBB = 5 V +−
υi
1) DC first 2) Then AC
VBE ON( ) = 0.7 V Lundstrom: 2019
1) DC analysis
15
+V
O= 5 V
−
β = 100
VCC = 10 V
RC = 5 kΩIC = 1mA
RBB = 430 kΩ
VBB = 5 V +−
VBE ON( ) = 0.7 V Lundstrom: 2019
2) AC analysis
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+υo
−
RC = 5 kΩic
RBB = 430 kΩυi
We need an ac small signal model for the BJT
Lundstrom: 2019
Recall: DC currents
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N P N
IC = β IB =α IE
IB = IC β
IE =β +1β
IC = ICα
What is the a.c. small signal model?
Lundstrom: 2019
ac collector current
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N P N
IC + ic
IB + ib
IC = ISeVBE VT → IC + ic = ISe
VBE+υbe( ) VT
IC + ic = ISeVBE VT eυbe VT
IC + ic = ISeVBE VT 1+υbe VT( )
ic = ISeVBE VT υbe VT( ) = IC υbe VT( )
ic =ICVT
υbe = gmυbe Ohm’s Law
Lundstrom: 2019
Summary: s.s. collector current
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N P N
ic
ib ic = gmυbe
gm = ICVT
+υbe
−
“transconductance”
Note that the ac model parameter, gm, depends on the dc bias current, IC. Lundstrom: 2019
s.s. base current
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N P N
ic
ib
ic = gmυbe
gm = ICVT
ib =icβ= gm
βυbe =
υbe
β gm= υbe
rπ
ib =υbe
rπ
gmrπ = β
+υbe
−
Lundstrom: 2019
s.s. eqv. circuit model: gm form
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icib
B
E
C
gm = IC VT gmrπ = β
+υbe
−E
C B
ib ic
rπ gmυπ
+υπ
−
“hybrid pi model”
s.s. eqv. circuit model: beta form
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gm = ICVT
gmrπ = βE
C B ib ic
rπ gmυπ
+υπ
−
gm = β rπ
ic =βrπυπ = βib
βib
ic = gmυπ
VCCS
CCCS
Lundstrom: 2019
s.s. eqv. circuit model: vce dependence
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E
C B ib ic
rπ gmυπ
+υπ
−
This model says that the collector current does not depend on the collector-emitter voltage.
Lundstrom: 2019
Output resistance
24 E
VCE
IC
IC = ISeqVBE kBT 1+
υCE
VA
⎛
⎝⎜⎞
⎠⎟
dIC
dVCE
= ISeqVBE kBT 1VA
⎛
⎝⎜⎞
⎠⎟=
′IC
VA
ic =
υce
r0
ro ≈
VA
IC
VA = “Early Voltage”
dIC
dVCE
≈icυceVBE1, IB1
Hybrid pi model
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gm = ICVT
gmrπ = βE
C B
ib ic
rπ gmυπ
+υπ
−
ro
+υce
−
ro =VA ICLundstrom: 2019
Small signal circuit
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+υo
−
RC = 5 kΩiC = IC + ic
RBB = 430 kΩυi
We now have an ac small signal model for the BJT
Lundstrom: 2019
Simple s.s. model
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gm = ICVT
gmrπ = β
E
C B ib ic
rπ+υπ
−
gmυπ
Lundstrom: 2019
ac analysis
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+υo
−
RC
ic
RBB
υi
C B ib
rπ+υπ
−
gmυπ
E
Same circuit
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+υo
− RC
ic RBBυs
C B
ib rπ+υπ
−
gmυπ
E
Lundstrom: 2019
ac analysis
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υo = −gmυπRC
υπ =rπ
rπ+ RBB
υs
υo = −rπ
rπ+ RBB
⎛
⎝⎜
⎞
⎠⎟ gmRCυs
υo
υs
= Aυs= −
rπ
rπ+ RBB
⎛
⎝⎜
⎞
⎠⎟ gmRC
β = 100
RC = 5 kΩIC = 1mA
RBB = 430 kΩ
gm = ICVT
= 1mA0.026 V
= 38.5 mS
gmrπ = β
rπ = β gm = 100 0.039 = 2.6 kΩ
Aυs= −1.2
Hybrid pi model with output resistance
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gm = ICVT
gmrπ = βE
C B
ib ic
rπ gmυπ
+υπ
−
ro
+υce
−
ro =VA ICLundstrom: 2019
Circuit with output resistance
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+υo
− RC
ic RBBυi
C B
ib rπ+υπ
−
gmυπ
E
ro
Lundstrom: 2019
Transistor parameters
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N P N
ic
ib
+υbe
−
β = 100
VBE ON( ) = 0.7 V
VA = 100 V
ro =VAIC
= 100 V1mA
= 100 kΩ
Lundstrom: 2019
2) AC analysis with output resistance
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υo
υi
= A = −rπ
rπ+ RBB
⎛
⎝⎜
⎞
⎠⎟ gmRC
Av = −1.2→ Av = −1.1
υo
υi
= A = −rπ
rπ+ RBB
⎛
⎝⎜
⎞
⎠⎟ gmRC || r0
5 kΩ ||100 kΩ = 4.76 kΩ
(without ro)
(with ro)
Lundstrom: 2019
Hybrid pi model of BJT
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gm = ICVT
gmrπ = βE
C B
ib ic
rπ gmυπ
+υπ
−
ro
+υce
−
ro =VA ICLundstrom: 2019
Question: What does the small signal model for a PNP transistor like?
Summary
36 Lundstrom: 2019
The small signal model of a BJT consists of two resistors and one voltage-controlled current course. The values of the ac model parameters are determined by the dc bias current.
The DC bias circuit places the operating point in the portion of the Voltage Transfer Characteristics where the output voltage changes rapidly with input voltage.
Circuit analysis consists of two steps: 1) dc analysis to determine the OP, and 2) ac small signal analysis using the ac circuit model.
Transistor Amplifiers
Lundstrom: 2019 37
1) Voltage transfer characteristic (VTC) 2) Small signal model for BJT 3) Small signal analysis